A) \[\frac{1}{4}\]
B) \[\frac{1}{2\sqrt{2}}\]
C) \[\frac{1}{2}\]
D) \[\frac{1}{\sqrt{2}}\]
Correct Answer: B
Solution :
[b] \[\because \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{\pi }{8}+\frac{3\pi }{8}=\frac{\pi }{2}\] \[{{\cos }^{3}}\frac{\pi }{8}.\cos \frac{3\pi }{8}+{{\sin }^{3}}\frac{\pi }{8}.\sin \frac{3\pi }{8}\] \[={{\cos }^{3}}\frac{\pi }{8}.\sin \frac{\pi }{8}+{{\sin }^{3}}\frac{\pi }{8}.\cos \frac{\pi }{8}\] \[=\sin \frac{\pi }{8}.\cos \frac{\pi }{8}\] \[\left( \because \,\,\,\,\,\,\,\sin \theta .\cos \theta =\frac{1}{2}\sin 2\theta \right)\] \[=\frac{1}{2}\sin \frac{\pi }{4}\] \[=\frac{1}{2\sqrt{2}}\]
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